\documentclass{ctexart}

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\title{2.7.1 Theoretical questions}


\author{邵柯欣 \\ 专业：信息与计算科学 \\ 学号：3200103310}

\begin{document}
\maketitle
\section{$\uppercase\expandafter{\romannumeral1}$}
解:
\begin{equation}
  f(x)-p_1(f;x) = \frac{f''(\xi(x))}{2}(x-x_0)(x-x_1).
  \label{eq::01}
\end{equation}
$f(x) = \frac{1}{x}, x_0 = 1, x_1 = 2$.\\
$p_1(f;x) = a_1*x+a_0, f(x_0) = 1, f(x_1) = 0.5, f'(x) = -\frac{1}{x^2}, f''(x) = \frac{2}{x^3}.$\\
因为 $p_1(x_0) = f(x_0)$ and $p_1(x_1) = f(x_1)$, $p_1(f;x) = -\frac{x}{2}+\frac{3}{2}$.\\
所以 $f(x)-p_1(f;x) = \frac{1}{x}+\frac{x}{2}-\frac{3}{2} = \frac{(x-1)(x-2)}{\xi(x)^3}$,\\
i.e. $\xi(x) = \sqrt[3]{2*x}$.\\
$max\{\xi(x)\} = \xi(x1) = \sqrt[3]{4}$\\
$min\{\xi(x)\} = \xi(x0) = \sqrt[3]{2}$\\
$max\{f''(\xi(x))\} = f''(min\{\xi(x)\}) = \frac{2}{(\sqrt[3]{2})^3} = 1$\\

\section{$\uppercase\expandafter{\romannumeral2}$}
解：\\
$f(x_i) = f_i, i \in {0, 1, ..., n}; f'(x_i) = f[x_i, x_i] = 0, i \in {0, 1, ..., n-1}.$\\
\begin{align}
  p(x) & = \sum\limits_{i=1}^{n-1}\pi_i^2(x) * \{ f[x_0, x_0, ..., x_{i-1}, x_{i-1}, x_i] + f[x_0, x_0, ..., x_i, x_i] * (x - x_i) \} \notag\\
  & + \pi_n^2(x) * f[x_0, x_0, ..., x_{i-1}, x_{i-1}, x_n] \notag 
\end{align}
为所求函数。


\section{$\uppercase\expandafter{\romannumeral3}$}
$f(x) = e^x$\\
用数学归纳法证明：
\begin{equation}
  f[t, t+1, ..., t+n] = \frac{(e-1)^n}{n!}*e^t\\
  \label{eq::02}
\end{equation}

证明：\\
当$n = 1$时，$f[t, t+1] = \frac{f[t+1]-f[t]}{(t+1)-t} = (e-1)*e^t$，(\ref{eq::02})成立;\\
假设当$n = k$时(\ref{eq::02})成立;\\
当$n = k+1$时，\\
\begin{align}
  f[t, t+1, ..., t+k, t+k+1] &= \frac{f[t+1, t+1, ..., t+k+1]-f[t, t+1, ..., t+k]}{t+k+1 - t} \notag\\
  &= \frac{\frac{(e-1)^k}{k!}*e^{t+1} - \frac{(e-1)^k}{k!}*e^{t}}{k+1}\notag\\
  &= \frac{(e-1)^{k+1}}{(k+1)!}*e^t \notag
\end{align}

所以(\ref{eq::02})对任意的$n = k+1$成立。\\
综上所述，(\ref{eq::02})对任意的$n \in N$成立。

\section{$\uppercase\expandafter{\romannumeral4}$}
解：
$f(0) = 5, f(1) = 3, f(3) = 5, f(4) = 12.$\\
$x_0 = 0, x_1 = 1, x_2 = 3, x_3 = 4; f(x_0) = 5, f(x_1) = 3, f(x_2) = 5, f(x_3) = 12.$
use Newton formula to obtain $p_3(f;x)$\\
$f[x_0] = 5, f[x_1] = 3, f[x_2] = 5, f[x_3] = 12;$\\
$f[x_0, x_1] = -2, f[x_1, x_2] = 1, f[x_2, x_3] = 7;$\\
$f[x_0, x_1, x_2] = 1, f[x_1, x_2, x_3] = 2;$\\
$f[x_0, x_1, x_2, x_3] = 0.25.$\\
所以，
\begin{align}
  p_3(f; x) &= f[x_0] + f[x_0, x_1]*(x-x_0) + f[x_0, x_1, x_2]*(x-x_0)*(x-x_1)\notag\\
  &+ f[x_0, x_1, x_2, x_3]*(x-x_0)*(x-x_1)*(x-x_2) \notag\\
  &= \frac{x^3}{4} - \frac{9*x}{4} + 5 \notag
\end{align}

\section{$\uppercase\expandafter{\romannumeral5}$}
解：
$f(x) = x^7$\\
$f[0] = 0, f[1] = 1, f[2] = 128;$\\
$f[0, 1] = 1, f[1, 1] = 7, f[1, 2] = 127, f[2, 2] = 448;$\\
$f[0, 1, 1] = 6, f[1, 1, 1] = 21, f[1, 1, 2] = 120, f[1, 2, 2] = 321;$\\
$f[0, 1, 1, 1] = 15, f[1, 1, 1, 2] = 99, f[1, 1, 2, 2] = 201;$\\
$f[0, 1, 1, 1, 2] = 42, f[1, 1, 1, 2, 2] = 102;$\\
$f[0, 1, 1, 1, 2, 2] = 30.$\\

$f^{(5)}(\xi) = 2520*\xi^2 = f[0, 1, 1, 1, 2, 2] = 30.$\\
所以，$\xi = \sqrt{84}$。

\section{$\uppercase\expandafter{\romannumeral6}$}
解：
$f(0) = 1, f(1) = 2, f'(1) = -1, f(3) = f'(3) = 0.$\\
$f[0] =1, f[1] = 2, f[3] = 0;$\\
$f[0, 1] = 1, f[1, 1] = -1, f[1, 3] = -1, f[3, 3] = 0;$\\
$f[0, 1, 1] = -2, f[1, 1, 3] = 0, f[1, 3, 3] = \frac{1}{2};$\\
$f[0, 1, 1, 3] = \frac{2}{3}, f[1, 1, 3, 3] = \frac{1}{4};$\\
$f[0, 1, 1, 3, 3] = -\frac{5}{36}.$\\
\begin{align}
  f(2) &= f[0] + f[0, 1]*(2 - 0) + f[0, 1, 1]*(2-0)*(2-1)\notag\\
  &+ f[0, 1, 1, 3]*(2-0)*(2-1)*(2-1) \notag\\
  &+ f[0, 1, 1, 3, 3]*(2-0)(2-1)*(2-1)*(2-3) \notag\\
  &= \frac{11}{18} \notag
\end{align}

$error(x) = \frac{f^{(5)}(x)}{5!}*(x-0)*(x-1)*(x-1)*(x-3)*(x-3)$.\\
$error(2) \le \frac{M}{60}$.

\section{$\uppercase\expandafter{\romannumeral7}$}
\begin{equation}
  \bigtriangleup^k f(x) = k!*h^k*f[x_0, x_1, ..., x_k],
  \label{eq::03}
\end{equation}
\begin{equation}
  \bigtriangledown^k f(x) = k!*h^k*f[x_0, x_{-1}, ..., x_{-k}],
  \label{eq::04}
\end{equation}

数学归纳法：\\
当$n=1$时，$\bigtriangleup f(x) = f(x+h)-f(x) = 1!*h^1f[x_0, x_1]$，(\ref{eq::03})成立。\\
假设当$n = k$时，(\ref{eq::03})成立。\\
当$n = k+1$时，\\
\begin{align}
  \bigtriangleup^{k+1} f(x) = &\bigtriangleup \bigtriangleup^k f(x) \notag \\
  & = \bigtriangleup^k f(x+h) - \bigtriangleup^k f(x) \notag \\
  & = k!*h^k*f[x_1, x_2, ..., x_k, x_{k+1}] - k!*h^k*f[x_0, x_1, ..., x_k] \notag \\
  & = k!*h^k*(x_{k+1}-x_0)*\frac{f[x_1, x_2, ..., x_{k+1}] - f[x_0, x_1, ..., x_k]}{x_{k+1}-x_0} \notag \\
  & = (k+1)!*h^{k+1}*f[x_0, x_1, ..., x_{k+1}] \notag
\end{align}
(\ref{eq::03})成立，对对任意的$n = k$。\\
综上所述，(\ref{eq::03})成立，对任意的$n \in N$。\\
同理可证得(\ref{eq::04})成立。

\section{$\uppercase\expandafter{\romannumeral8}$}
\begin{equation}
  \dfrac{\partial}{\partial x_0}f[x_0, x_1, ..., x_n] = f[x_0, x_0, x_1, ..., x_n]
  \label{eq::05}
\end{equation}
当$n=1$时，$\dfrac{\partial}{\partial x_0}f[x_0, x_1] =\dfrac{\partial}{\partial x_0} \dfrac{f[x_1]-f[x_0]}{x_1-x_0} = \dfrac{(f[x_1]-f[x_0])-f[x_0,x_0]*(x_1-x_0)}{(x_1-x_0)^2}= f[x_0, x_0, x_1]$，(\ref{eq::05})成立。\\
假设当$n = k$时，(\ref{eq::05})成立。即，$\dfrac{\partial}{\partial x_0}f[x_0, x_1, ..., x_k] = f[x_0, x_0, x_1, ..., x_k]$。\\
当$n = k+1$时，\\
\begin{align}
  \dfrac{\partial}{\partial x_0}f[x_0, x_1, ..., x_{k+1}] &= \dfrac{\partial}{\partial x_0} \dfrac{f[x_1, x_2, ...,x_{k+1}]-f[x_0, x_1, ..., x_k]}{x_{k+1}-x_0} \notag\\
  &= \dfrac{(f[x_1, x_2, ..., x_{k+1}]-f[x_0, x_1, ...,x_k])-f[x_0, x_1, ..., x_k]*(x_{k+1}-x_0)}{(x_{k+1}-x_0)^2} \notag \\
  &= f[x_0, x_0, x_1, ..., x_n] \notag
\end{align}

(\ref{eq::05})成立，对对任意的$n = k$。\\
综上所述，(\ref{eq::05})成立，对任意的$n \in N$。

\section{$\uppercase\expandafter{\romannumeral9}$}
解：\\
对 $n \in N^+$, \\
$min max_{x \in [a,b]}|a_0*x^n + a_1*x^{n-1} + ...+a_n| = min max_{x \in [a,b]}|a_0*(x-x_0)^n + C|$.\\
其中C是一个常数, 并且 $x_0 = \frac{a+b}{2}$.\\
所以 $min max_{x \in [a,b]}|a_0*x^n + a_1*x^{n-1} + ...+a_n| = \dfrac{a_0}{2}*(\dfrac{a-b}{2})^n$.

\section{$\uppercase\expandafter{\romannumeral10}$}
$\forall p \in \mathbb{P}^a_n$, ${\|\hat{p_n} \|}_{\infty} \le {\|p\|}_{\infty}$. 
\section{$\uppercase\expandafter{\romannumeral11}$}
Lemma2.50
\begin{align}
  \forall k = 0,1,...,n, \forall t \in (0,1), b_{n,k}(t) &> 0 \notag \\
  \sum\limits^n_{k=0}b_{n,k}(t) &= 1, \notag \\
  \sum\limits^n_{k=0}k*b_{n,k}(t) &= n*t, \notag \\
  \sum\limits^n_{k=0}(k-nt)^2*b_{n,k}(t) &= n*t*(1-t). \notag 
\end{align}

\end{document}
